Solution for the following limit problem

Question

I want to solve the following limit problem $$\lim_{x \rightarrow \infty} \bigg[ (x + a)\log \big( \frac{x+a}{x+b} \big) \bigg] $$

A small simulation with a = 5 and b = 2 leads to a result of 3 which I think is correct and the right answer should be (a - b), somehow I am unable to prove it.

Here is the code to reproduce the result

def f(alpha):
    return (5 + alpha) * np.log((5 + alpha) / (2 + alpha))
alpha_list = np.linspace(10,1000)
plt.plot(alpha,[f(a) for a in alpha_list])

yields the following plot,

Thanks in advance.

score 9 · Accepted Answer · answered Apr 12 '21 at 19:57

9

Use the limit definition of $e$

$$-(x+a)\log\left(\frac{x+b}{x+a}\right) = -\log\left(1+\frac{b-a}{x+a}\right)^{x+a}$$ $$\longrightarrow -\log e^{b-a} = a-b$$

answered Apr 12 '21 at 19:57

Ninad Munshi

34,407

Elegant solution. – saulspatz Apr 12 '21 at 20:02
Really liked the solution!! thanks – Hirak Sarkar Apr 12 '21 at 20:04
Thanks for the kind words everyone! – Ninad Munshi Apr 12 '21 at 21:00
(+1) great way of thinking about it! – Henry Lee Apr 13 '21 at 10:16

score 4 · Answer 2 · answered Apr 12 '21 at 20:01

Lets think logically about what is going on here, as $x\to\infty$: $$\log\left(\frac{x+a}{x+b}\right)\approx\log\left(\frac xx\right)=\log(1)=0$$ as $x$ will be much larger than $a,b$. Now in the second part: $$x+a\to\infty$$ and this is pretty obvious. So, to put it crudely we have a $0\times\infty$ situation. If you want to use L'Hopital's rule you can do the following: $$(x+a)\log\left(\frac{x+a}{x+b}\right)=\frac{\log(x+a)-\log(x+b)}{\frac{1}{x+a}}$$ Now both the top and bottom of this fraction tend to zero, so we can say: $$\lim_{x\to\infty^+}\frac{\log(x+a)-\log(x+b)}{\frac{1}{x+a}}=\lim_{x\to\infty^+}\frac{\frac{1}{x+a}-\frac{1}{x+b}}{-\frac{1}{(x+a)^2}}$$ $$=\lim_{x\to\infty^+}\frac{(x+a)^2}{(x+b)}-(x+a)$$ which we can rewrite as: $$\lim_{x\to\infty^+}\frac{(x+a)\left[(x+a)-(x+b)\right]}{(x+b)}=\lim_{x\to\infty^+}\frac{(x+a)(a-b)}{(x+b)}=\lim_{x\to\infty^+}(a-b)=a-b$$ which is probably the easiest way to do it

score 2 · Answer 3 · answered Apr 12 '21 at 20:06

2

With $ y=\frac 1x $, it becomes

$$\lim_{y\to 0^+}\frac{1+ay}{y}\Bigl(\ln(1+ay)-\ln(1+by)\Bigr)=$$

$$\lim_{y\to 0^+}(1+ay)\Bigl(\frac{\ln(1+ay)}{ay}a-\frac{\ln(1+by)}{by}b\Bigr)=$$ $$1.(1.a-1.b)=a-b$$

answered Apr 12 '21 at 20:06

hamam_Abdallah

62,951

robjohn · Answer 4 · 2021-04-13T22:55:22.357

2

Note that $$ \begin{align} (x+a)\log\left(\frac{x+a}{x+b}\right) &=(x+a)\int_{x+b}^{x+a}\frac1t\,\mathrm{d}t\tag{1a}\\ &=(x+a)\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $\log(t)$ is the integral of $1/t$
$\text{(1b)}$: substitute $t\mapsto xt$

Next, we have the bounds $$ \overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+a/x}}^{\substack{\text{minimum of}\\\text{integrand}}}\le\int_{1+b/x}^{1+a/x}\frac1t\,\mathrm{d}t\le\overbrace{\ \ \frac{a-b}x\ \ }^{\substack{\text{width of}\\\text{domain}}}\overbrace{\frac1{1+b/x}}^{\substack{\text{maximum of}\\\text{integrand}}}\tag2 $$ Multiplying $(2)$ by $x+a$ and then applying $(1)$, we get $$ a-b\le(x+a)\log\left(\frac{x+a}{x+b}\right)\le(a-b)\frac{x+a}{x+b}\tag3 $$ Graphically, we can see the function squeezed between two functions whose limits are easier to compute:

The Squeeze Theorem then gives $$ \lim_{x\to\infty}(x+a)\log\left(\frac{x+a}{x+b}\right)=a-b\tag4 $$

edited Apr 13 '21 at 22:55

answered Apr 13 '21 at 10:12

robjohn

345,667

Very interesting, just wondering, is squeeze theorem true for all (real-valued) functions? – Hirak Sarkar Apr 13 '21 at 13:44
If $f_-(x)\le f(x)\le f_+(x)$ for all $x$ near $a$ and $\lim\limits_{x\to a}f_-(x)=\lim\limits_{x\to a}f_+(x)=L$, then $\lim\limits_{x\to a}f(x)=L$, where "near" $\infty$ means sufficiently large. – robjohn Apr 13 '21 at 15:04
Note that $\lim\limits_{x\to\infty}(a-b)=\lim\limits_{x\to\infty}\frac{x+a}{x+b}(a-b)=a-b$ – robjohn Apr 13 '21 at 15:08
I have added a link to the Squeeze Theorem. – robjohn Apr 13 '21 at 15:16
I have also added an illustration showing the "squeezing". – robjohn Apr 13 '21 at 22:55
Great explanation @robjohn, appreciate it. – Hirak Sarkar Apr 14 '21 at 00:24

score 1 · Answer 5 · answered Apr 12 '21 at 20:11

1

A straightforward solution uses the Taylor series for $\log(1+x)$.

$$\log\frac{x+a}{x+b}=\log\left(1+\frac{a-b}{x+b}\right)=\frac{a-b}{x+b}+O\left(\frac1{(x+b)^2}\right)$$ $$\lim_{x\to\infty}(x+a)\log\frac{x+a}{x+b}=\lim_{x\to\infty}(a-b)\frac{x+a}{x+b}=a-b$$

answered Apr 12 '21 at 20:11

saulspatz

53,131

Your last step is l'hopitals? – Hirak Sarkar Apr 12 '21 at 20:14
2

@HirakSarkar If you mean how to get $\lim_{x\to\infty}\frac{x+a}{x+b}=1$, you can use L'Hospital, or just divide numerator and denominator by $x$. – saulspatz Apr 12 '21 at 20:16

Solution for the following limit problem

5 Answers5